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\[\int\limits_{\sqrt{2}}^{2}\int\limits_{\sqrt{4-y^{2}}}^{y} dx dy\]

|dw:1353881320734:dw|i think this is what it looks like at least

remember that x=rcos(t)
y=rsin(t)
and r=sqrt(x^2+y^2), but thats really already given as 2

I think I understand how to get pi/4 < theta < pi/2
But finding the bounds for r is throwing me off.

well, since r only really changes from 0 to 2; the bounds for r are 0 to 2

it r changed according to some function of t
the r would be bound from 0 to f(t)

according to the answers, r is bounded by 2 < r < 2csc(theta)

|dw:1353881730467:dw|

but at pi/2 r can be 2 and rad(2)

y = 2
2sin(t) = 2
sin(t) = 1; when t=pi/2
y=sqrt(2)
2sin(t)=sqrt(2)
sin(t)=sqrt(2)/2; when t=pi/4

i might need to see the steps in your books to refresh me memory

I just dont understand where they cot the csc from.

your d parts in that png are dydx ... not dxdy

well, thats a problem...
the question has it as dx dy

I dont think dy dx even works. Wouldnt it be y=y y=rad(4-y^2)

yeah, so the png might be a typo

ok so ignoring the png, would the r just be 0 to 2?

from what i can tell yes;
|dw:1353883154486:dw|

but then i might have been thinking of this backwards

so, r goes from 2 up to the line; its not the area inside the circle

But doesnt the outer circle ask that the Y bounds are 2 and rad(2) not the X?

oh you flipped it

i positioned the graph in a convenient manner; a swaped the x and y axis ,,, yes

so, what we want is a function in ploar from for the line y=2

well, even though the answer key messed up the dxdy, but 2-(pi/2) seems to be the right answer.

it is; and youd be able to see that y=2 converts to 2csc(t) if i could remember how to do it :)

not sure if its coincidence or not; but that seems to work out as well :)